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    <div class="post-body" itemprop="articleBody"><h1 id="二分答案">二分答案</h1>
<p>在线性表章节已了解二分查找的基本原理，当一个问题的答案具有单调性时——即随着答案的增大或减小，判定条件的结果也呈现单调变化，比如答案越大条件越容易满足，或者答案越小条件越容易满足，可以通过二分查找，在过程中根据判定条件是否满足来调整二分的方向逼近解的过程。</p>
<span id="more"></span>
<h3 id="例一元三次方程求解">例：一元三次方程求解</h3>
<p>给定一个一元三次方程 <span class="math inline">\(ax^3 + bx^2 + cx + d
= 0\)</span>，其中系数 <span class="math inline">\(a, b, c, d\)</span>
均为实数，并已知该方程有三个不同的实根，这些根的范围在 <span
class="math inline">\(-100\)</span> 至 <span
class="math inline">\(100\)</span> 之间，且任意两根间的差的绝对值 <span
class="math inline">\(\geq
1\)</span>。请按从小到大的顺序在同一行输出这三个实根（精确到小数点后两位，并留有空格）。</p>
<p>输入：4个实数</p>
<figure class="highlight txt"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">1 -5 -4 20</span><br></pre></td></tr></table></figure>
<p>输出：3个保留2位小数的实根</p>
<figure class="highlight txt"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">-2.00 2.00 5.00</span><br></pre></td></tr></table></figure>
<p>题目给的关键条件是数据确保根的距离大于 <span
class="math inline">\(1\)</span>，也即相邻两个整数的左闭右开区间中至多 1
个解。如果某个这样的区间有解，则必然两端的 <span
class="math inline">\(f(x)\)</span> 不可能在 <span
class="math inline">\(x\)</span> 轴的同一侧，即 <span
class="math inline">\(f(x_{1})*f(x_{2}) \leq 0\)</span>。</p>
<p>思路：枚举每个长度为 1 的区间，判断两端是否不在 <span
class="math inline">\(x\)</span>
轴同侧，如果不在同侧，说明区间内有解，二分 <span
class="math inline">\(x\)</span> 的值，直到它的函数值为 0。</p>
<figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br></pre></td><td class="code"><pre><span class="line"><span class="meta">#<span class="keyword">include</span><span class="string">&lt;cstdio&gt;</span></span></span><br><span class="line"><span class="type">const</span> <span class="type">double</span> eps = <span class="number">1e-7</span>;</span><br><span class="line"><span class="function"><span class="type">double</span> <span class="title">f</span><span class="params">(<span class="type">double</span> x, <span class="type">double</span> a, <span class="type">double</span> b, <span class="type">double</span> c, <span class="type">double</span> d)</span> </span>&#123;</span><br><span class="line">    <span class="keyword">return</span> a * x * x * x + b * x * x + c * x + d;</span><br><span class="line">&#125;</span><br><span class="line"></span><br><span class="line"><span class="function"><span class="type">int</span> <span class="title">main</span><span class="params">()</span> </span>&#123;</span><br><span class="line">    <span class="type">double</span> a, b, c, d;</span><br><span class="line">    <span class="built_in">scanf</span>(<span class="string">&quot;%lf%lf%lf%lf&quot;</span>, &amp;a, &amp;b, &amp;c, &amp;d);</span><br><span class="line">    </span><br><span class="line">    <span class="comment">// 在[-100,100]范围内，相邻整数区间最多一个解</span></span><br><span class="line">    <span class="comment">// 遍历每个整数区间，用二分法找解</span></span><br><span class="line">    <span class="keyword">for</span>(<span class="type">int</span> i = <span class="number">-100</span>; i &lt;= <span class="number">100</span>; i ++) &#123;</span><br><span class="line">        <span class="type">double</span> l = i, r = i + <span class="number">1</span>;</span><br><span class="line">        <span class="comment">// 区间两端异号说明有解</span></span><br><span class="line">        <span class="type">double</span> fl = <span class="built_in">f</span>(l, a, b, c, d), fr = <span class="built_in">f</span>(r, a, b, c, d);</span><br><span class="line">        <span class="keyword">if</span>(-eps &lt; fl &amp;&amp; fl &lt; eps) &#123;</span><br><span class="line">            <span class="built_in">printf</span>(<span class="string">&quot;%.2f &quot;</span>, l); <span class="comment">// 左端点是解</span></span><br><span class="line">        &#125; <span class="keyword">else</span> <span class="keyword">if</span>(-eps &lt; fr &amp;&amp; fr &lt; eps) &#123;</span><br><span class="line">            <span class="keyword">continue</span>;           <span class="comment">// 右端点是解，考虑左闭右开区间，忽略</span></span><br><span class="line">        &#125; <span class="keyword">else</span> <span class="keyword">if</span>(fl * fr &lt; eps) &#123;</span><br><span class="line">            <span class="comment">// l 和 r 之间存在解，二分找解</span></span><br><span class="line">            <span class="keyword">for</span>(<span class="type">int</span> j = <span class="number">0</span>; j &lt; <span class="number">100</span>; j ++) &#123;</span><br><span class="line">                <span class="type">double</span> mid = (l + r) * <span class="number">0.5</span>;</span><br><span class="line">                <span class="type">double</span> fm = <span class="built_in">f</span>(mid, a, b, c, d);</span><br><span class="line">                <span class="keyword">if</span>(fm * fl &lt; eps) r = mid;</span><br><span class="line">                <span class="keyword">else</span> l = mid;</span><br><span class="line">            &#125;</span><br><span class="line">            <span class="built_in">printf</span>(<span class="string">&quot;%.2f &quot;</span>, l);</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="built_in">printf</span>(<span class="string">&quot;\n&quot;</span>);</span><br><span class="line">    <span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>
<blockquote>
<p>知识点： 浮点精度冗余 <code>eps</code>。
浮点类型在计算机底层也是由有限位的二进制<code>01</code>表达的，就必然有精度限制，不能无限精确地表达任何数字，如果发生乘除法，这种精度会进一步丢失，当判断两个浮点数是否相等、一个浮点数大于0还是小于0的时候，用一个极小的浮点数作为精度冗余，比如对浮点数
<code>fl</code>， <code>if(fl == 0)</code>
就是不安全的，它在丧失一定精度后，可能是
<code>0.0000312141</code>，也可能是<code>-0.0000012131</code>，
<code>eps</code>
是我们对特定问题期望的一个“可接受”精度，<code>const double eps=1e-7</code>
就表示在 <code>10^&#123;-7&#125;</code>（<code>0.0000001</code>）
范围内的误差都认为相等，<code>1e-7</code>
是算法题比较常用的一个精度，视具体问题，也有时会用
<code>1e-6</code>或<code>1e-8</code>。</p>
</blockquote>
<blockquote>
<p>知识点： 浮点数二分的时候，<code>while(r - l &gt; eps)</code>
是循环条件的常见写法，但一些刁钻的题可能精度不好把握，<code>for(int j = 0; j &lt; 100; j ++)</code>
直接二分<code>100</code>次，基本什么精度都能达到了，在分析好算法复杂度满足题意的情况下，这样写更加稳定。</p>
</blockquote>
<h3 id="例数列划分">例：数列划分</h3>
<p>给一个有 <span class="math inline">\(n\)</span>
个正整数的序列，连续地将其分为互不重叠的 <span
class="math inline">\(k\)</span>
份，即每一份都是该序列的一个子串，不能调换顺序。</p>
<p>求一个划分方案，使数字之和最大的那一份，其和在所有划分方案里最小。</p>
<p>输入</p>
<figure class="highlight txt"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br></pre></td><td class="code"><pre><span class="line">9 3</span><br><span class="line">10 20 30 40 50 60 70 80 90</span><br></pre></td></tr></table></figure>
<p>输出</p>
<figure class="highlight txt"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">10 20 30 40 50 / 60 70 / 80 90</span><br></pre></td></tr></table></figure>
<p>二分答案常常伴随着<strong>贪心</strong>策略——二分可能的答案数值，在验证判定条件时，往往会需要一个复杂度较低的策略，当判定条件支持贪心时，就有更大的可能让二分答案的复杂度低于其它方案。</p>
<p>当指定一个“数字之和最大的那一份”的限制时，实际就是尝试构造一个划分，让每一份都不超过这个限制，如果无法实现，则判定为“否”，如果能实现，则判定为“是”，限制值与判定关系是单调的——值够大就一定能成功划分，值小到一定程度就一定无法划分，那么能否完成划分的临界点，就是本题答案——最大那一份最小的划分。</p>
<p>划分的判定采用贪心策略，因为划分并不改变顺序，那就让每一份尽可能的靠近限制值，直到装不下，再开始划分下一份。</p>
<figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br></pre></td><td class="code"><pre><span class="line"><span class="meta">#<span class="keyword">include</span> <span class="string">&lt;cstdio&gt;</span></span></span><br><span class="line"><span class="meta">#<span class="keyword">include</span> <span class="string">&lt;cstdlib&gt;</span></span></span><br><span class="line"><span class="meta">#<span class="keyword">include</span> <span class="string">&lt;cstring&gt;</span></span></span><br><span class="line"></span><br><span class="line"><span class="meta">#<span class="keyword">include</span> <span class="string">&lt;algorithm&gt;</span></span></span><br><span class="line"><span class="meta">#<span class="keyword">include</span> <span class="string">&lt;cmath&gt;</span></span></span><br><span class="line"><span class="meta">#<span class="keyword">include</span> <span class="string">&lt;complex&gt;</span></span></span><br><span class="line"><span class="meta">#<span class="keyword">include</span> <span class="string">&lt;vector&gt;</span></span></span><br><span class="line"></span><br><span class="line"><span class="type">const</span> <span class="type">int</span> maxn = <span class="number">1e5</span> + <span class="number">10</span>;</span><br><span class="line"><span class="type">int</span> n, k, a[maxn];</span><br><span class="line"><span class="type">bool</span> split[maxn];</span><br><span class="line"><span class="function"><span class="type">bool</span> <span class="title">Judge</span><span class="params">(<span class="type">int</span> mid)</span> </span>&#123;</span><br><span class="line">    <span class="type">int</span> cnt = <span class="number">1</span>, tmpsum = <span class="number">0</span>;</span><br><span class="line">    <span class="keyword">for</span> (<span class="type">int</span> i = <span class="number">0</span>; i &lt; n; i++) &#123;</span><br><span class="line">        <span class="keyword">if</span> (a[i] &gt; mid) <span class="keyword">return</span> <span class="literal">false</span>;</span><br><span class="line">        <span class="keyword">if</span> (tmpsum + a[i] &gt; mid) tmpsum = <span class="number">0</span>, cnt++;</span><br><span class="line">        tmpsum += a[i];</span><br><span class="line">        <span class="keyword">if</span> (cnt &gt; k) <span class="keyword">return</span> <span class="literal">false</span>;</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="keyword">return</span> <span class="literal">true</span>;</span><br><span class="line">&#125;</span><br><span class="line"><span class="function"><span class="type">int</span> <span class="title">main</span><span class="params">()</span> </span>&#123;</span><br><span class="line">    <span class="keyword">while</span> (<span class="built_in">scanf</span>(<span class="string">&quot;%d%d&quot;</span>, &amp;n, &amp;k) != EOF) &#123;</span><br><span class="line">        <span class="keyword">for</span> (<span class="type">int</span> i = <span class="number">0</span>; i &lt; n; i++) <span class="built_in">scanf</span>(<span class="string">&quot;%d&quot;</span>, &amp;a[i]);</span><br><span class="line">        <span class="type">int</span> left = <span class="number">0</span>, right = <span class="number">1e9</span>, mid;</span><br><span class="line">        <span class="keyword">while</span> (left &lt; right) &#123;</span><br><span class="line">            mid = left + right &gt;&gt; <span class="number">1</span>;</span><br><span class="line">            <span class="keyword">if</span> (<span class="built_in">Judge</span>(mid))</span><br><span class="line">                right = mid;</span><br><span class="line">            <span class="keyword">else</span></span><br><span class="line">                left = mid + <span class="number">1</span>;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="type">int</span> tmpsum = <span class="number">0</span>, cnt = <span class="number">0</span>;</span><br><span class="line">        <span class="built_in">memset</span>(split, <span class="number">0</span>, <span class="built_in">sizeof</span>(split));</span><br><span class="line">        <span class="keyword">for</span> (<span class="type">int</span> i = n - <span class="number">1</span>; i &gt;= <span class="number">0</span>; i--) &#123;</span><br><span class="line">            <span class="keyword">if</span> (tmpsum + a[i] &gt; left || i + <span class="number">1</span> &lt; k - cnt)</span><br><span class="line">                tmpsum = <span class="number">0</span>, split[i] = <span class="literal">true</span>, cnt++;</span><br><span class="line">            tmpsum += a[i];</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">for</span> (<span class="type">int</span> i = <span class="number">0</span>; i &lt; n; i++) &#123;</span><br><span class="line">            <span class="built_in">printf</span>(<span class="string">&quot; %d&quot;</span> + !i, a[i]);</span><br><span class="line">            <span class="keyword">if</span> (split[i]) <span class="built_in">printf</span>(<span class="string">&quot; /&quot;</span>);</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="built_in">printf</span>(<span class="string">&quot;\n&quot;</span>);</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>

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